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Question #44341

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the local geometry around the central heteroatom (N or O) for the following molecules in the gas phase at $298\text{ K}$ : (I) $\ce{SiH3NCO}$ (II) $\ce{CH3NCO}$ (III) $\ce{SiH3NCS}$ (IV) $\ce{N(SiH3)3}$ (V) $\ce{O(SiH3)2}$ Assign a structural code to each molecule: '1' for Linear, '2' for Trigonal Planar, and '3' for Bent/Pyramidal. Which of the following sequences correctly represents the geometries of (I) through (V)?

Options

A.
1, 3, 3, 2, 3
B.
1, 3, 1, 2, 3
C.
3, 3, 3, 2, 1
D.
1, 1, 3, 2, 3

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

VSEPR Lone Pair Positioning|Synergic Bonding in Carbonyls

Why This Question Is Tricky

Common Mistakes

This question tests the nuanced limits of $p\pi-d\pi$ back-bonding, a common trap for advanced students.

While students correctly identify back-bonding in $\ce{N(SiH3)3}$ (making it planar) and $\ce{SiH3NCO}$ (making it linear), they often overgeneralize this rule to all silicon-heteroatom systems.

The critical exception is $\ce{SiH3NCS}$ , where the lower electronegativity of sulfur reduces the efficiency of back-donation, failing to force linearity and leaving the molecule bent.

Similarly, students often mistakenly assume $\ce{O(SiH3)2}$ is linear due to back-bonding, ignoring that the presence of a second lone pair keeps the geometry bent despite significant angle widening.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Bonding and Molecular Structure - Class 11 Part 1
Concept	VSEPR Lone Pair Positioning
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free